C4 June 2013 (R) Q8
8. In an experiment testing solid rocket fuel, some fuel is burned and the waste products are collected. Throughout the experiment the sum of the masses of the unburned fuel and waste products remains constant.
Let \(x\) be the mass of waste products, in kg, at time \(t\) minutes after the start of the experiment. It is known that at time \(t\) minutes, the rate of increase of the mass of waste products, in kg per minute, is \(k\) times the mass of unburned fuel remaining, where \(k\) is a positive constant.
The differential equation connecting \(x\) and \(t\) may be written in the form \[\frac{\mathrm{d}x}{\mathrm{d}t} = k(M - x), \text{ where } M \text{ is a constant.}\]
Given that initially the mass of waste products is zero,
Given also that \(x = \dfrac{1}{2}M\) when \(t = \ln 4\),
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = k(M - x)\), where \(M\) is a constant | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) is the rate of increase of the mass of waste products. Any one correct explanation. | B1 |
| \(M\) is the total mass of unburned fuel and waste fuel (or the initial mass of unburned fuel) Both explanations are correct. | B1 |
| (2) |
Notes
B1: At least one explanation correct.
B1: Both explanations are correct.
\(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) is the rate of increase of the mass of waste products.
or the rate of change of the mass of waste products.
\(M\) is the total mass of unburned fuel and waste fuel
or the initial mass of unburned fuel
or the total mass of rocket fuel and waste fuel
or the initial mass of rocket fuel
or the initial mass of fuel
or the total mass of waste and unburned products.
| Scheme | Marks | ||||||||
|---|---|---|---|---|---|---|---|---|---|
| \(\displaystyle\int \dfrac{1}{M - x}\,\mathrm{d}x = \displaystyle\int k\,\mathrm{d}t\) or \(\displaystyle\int \dfrac{1}{k(M - x)}\,\mathrm{d}x = \displaystyle\int \mathrm{d}t\) | B1 | ||||||||
| \(-\ln(M - x) = kt\ \{+ c\}\) or \(-\dfrac{1}{k}\ln(M - x) = t\ \{+ c\}\) See notes | M1 A1 | ||||||||
| \(\{t = 0,\ x = 0 \Rightarrow\}\ -\ln(M - 0) = k(0) + c\) See notes | M1 | ||||||||
| \(c = -\ln M \Rightarrow -\ln(M - x) = kt - \ln M\) | |||||||||
| ddM1 A1 * cso | ||||||||
| (6) | |||||||||
Notes
B1: Separates variables as shown. \(\mathrm{d}x\) and \(\mathrm{d}t\) should be in the correct positions, though this mark can be implied by later working. Ignore the integral signs.
M1: Both \(\pm\lambda\ln(M - x)\) or \(\pm\lambda\ln(x - M)\) and \(\pm\mu t\) where \(\lambda\) and \(\mu\) are any constants.
A1: For \(-\ln(M - x) = kt\) or \(-\ln(x - M) = kt\) or \(-\dfrac{1}{k}\ln(M - x) = t\) or \(-\dfrac{1}{k}\ln(x - M) = t\)
or \(-\dfrac{1}{k}\ln(kM - kx) = t\) or \(-\dfrac{1}{k}\ln(kx - kM) = t\)
Note: \(+c\) is not needed for this mark.
IMPORTANT: \(+\,c\) can be on either side of their equation for the 1st A1 mark.
M1: Substitutes \(t = 0\) AND \(x = 0\) in an integrated or changed equation containing \(c\) (or \(A\) or \(\ln A\), etc.)
Note that this mark can be implied by the correct value of \(c\).
ddM1: Uses their value of \(c\) which must be a \(\ln\) term, and uses fully correct method to eliminate their logarithms. Note: This mark is dependent on both previous method marks being awarded.
A1: \(x = M - M\mathrm{e}^{-kt}\) or \(x = M(1 - \mathrm{e}^{-kt})\) or \(x = \dfrac{M(\mathrm{e}^{kt} - 1)}{\mathrm{e}^{kt}}\) or equivalent where \(x\) is the subject.
Note: Please check their working as incorrect working can lead to a correct answer.
Note: \(\left\{\dfrac{\mathrm{d}x}{\mathrm{d}t} = k(M - x) \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1}{kM - kx} \Rightarrow\right\}\ t = -\dfrac{1}{k}\ln(kM - kx)\ \{+ c\}\) is B1(Implied) M1A1.
(corrected from the printed mark scheme: this note is printed with \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{kM - kx}\) and \(x = -\dfrac{1}{k}\ln(kM - kx)\ \{+ c\}\); it should be \(\dfrac{\mathrm{d}t}{\mathrm{d}x}\) and \(t\).)
Aliter 8. (b) Way 2
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \dfrac{1}{M - x}\,\mathrm{d}x = \displaystyle\int k\,\mathrm{d}t\) | B1 |
| \(-\ln(M - x) = kt\ \{+ c\}\) See notes | M1 A1 |
| \(\ln(M - x) = -kt + c\) \(M - x = A\mathrm{e}^{-kt}\) | |
| \(\{t = 0,\ x = 0 \Rightarrow\}\ M - 0 = A\mathrm{e}^{-k(0)}\) \(\Rightarrow M = A\) | M1 |
| \(M - x = M\mathrm{e}^{-kt}\) | ddM1 |
| So, \(x = M - M\mathrm{e}^{-kt}\) | A1 |
| (6) |
B1M1A1: Mark as in the original scheme.
M1: Substitutes \(t = 0\) AND \(x = 0\) in an integrated equation containing their constant of integration which could be \(c\) or \(A\). Note that this mark can be implied by the correct value of \(c\) or \(A\).
ddM1: Uses a fully correct method to eliminate their logarithms and writes down an equation containing their evaluated constant of integration.
Note: This mark is dependent on both previous method marks being awarded.
Note: \(\ln(M - x) = -kt + c\) leading to \(\ln(M - x) = \mathrm{e}^{-kt} + \mathrm{e}^c\) or \(\ln(M - x) = \mathrm{e}^{-kt} + A\) would be dddM0.
A1: Same as the original scheme.
Aliter 8. (b) Way 3
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^x \dfrac{1}{M - x}\,\mathrm{d}x = \displaystyle\int_0^t k\ \mathrm{d}t\) | B1 |
| \(\left[-\ln(M - x)\right]_0^x = \left[kt\right]_0^t\) | M1 A1 |
| \(-\ln(M - x) - (-\ln M) = kt\) \(-\ln(M - x) + \ln M = kt\) and then follows the original scheme. Applies limits of | M1 |
B1M1A1: Mark as in the original scheme (ignoring the limits).
ddM1: Applies limits 0 and \(x\) on their integrated LHS and limits of 0 and \(t\).
M1A1: Same as the original scheme.
Aliter 8. (b) Way 4
| Scheme | Marks | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\displaystyle\int \dfrac{1}{M - x}\,\mathrm{d}x = \displaystyle\int k\,\mathrm{d}t\ \left\{\Rightarrow \displaystyle\int \dfrac{-1}{x - M}\,\mathrm{d}x = \displaystyle\int k\,\mathrm{d}t\right\}\) | B1 | ||||||||||||
| \(-\ln|x - M| = kt + c\) Modulus not required for 1st A1. | M1 A1 | ||||||||||||
| \(\{t = 0,\ x = 0 \Rightarrow\}\ -\ln|0 - M| = k(0) + c\) Modulus not required here! | M1 | ||||||||||||
| \(\Rightarrow c = -\ln M \Rightarrow -\ln|x - M| = kt - \ln M\) | |||||||||||||
Understanding of modulus is required here! | ddM1 A1 * cso | ||||||||||||
| (6) | |||||||||||||
B1: Mark as in the original scheme.
M1A1M1: Mark as in the original scheme ignoring the modulus.
ddM1: Mark as in the original scheme AND the candidate must demonstrate that they have converted \(\ln|x - M|\) to \(\ln(M - x)\) in their working.
Note: This mark is dependent on both the previous method marks being awarded.
A1: Mark as in the original scheme.
Aliter 8. (b) Way 5
Use of an integrating factor (I.F.)
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = k(M - x) \Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} + kx = kM\) | |
| \(\text{I.F.} = \mathrm{e}^{kt}\) | B1 |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{kt}x\right) = kM\mathrm{e}^{kt}\), | |
| \(\mathrm{e}^{kt}x = M\mathrm{e}^{kt} + c\) | M1A1 |
| \(x = M + c\mathrm{e}^{-kt}\) | |
| \(\{t = 0,\ x = 0 \Rightarrow\}\ 0 = M + c\mathrm{e}^{-k(0)}\) | M1 |
| \(\Rightarrow c = -M\) | |
| \(x = M - M\mathrm{e}^{-kt}\) | ddM1A1 |
| Scheme | Marks |
|---|---|
| \(\left\{x = \dfrac{1}{2}M,\ t = \ln 4 \Rightarrow\right\}\ \dfrac{1}{2}M = M(1 - \mathrm{e}^{-k\ln 4})\) | M1 |
| \(\Rightarrow \dfrac{1}{2} = 1 - \mathrm{e}^{-k\ln 4} \Rightarrow \mathrm{e}^{-k\ln 4} = \dfrac{1}{2} \Rightarrow -k\ln 4 = -\ln 2\) | |
| So \(k = \dfrac{1}{2}\) | A1 |
| \(x = M\left(1 - \mathrm{e}^{-\frac{1}{2}\ln 9}\right)\) | dM1 |
| \(x = \dfrac{2}{3}M\) \(x = \dfrac{2}{3}M\) | A1 cso |
| (4) | |
| (12 marks) |
Notes
M1: Substitutes \(x = \dfrac{1}{2}M\) and \(t = \ln 4\) into one of their earlier equations connecting \(x\) and \(t\).
A1: \(k = \dfrac{1}{2}\), which can be an un-simplified equivalent numerical value. i.e. \(k = \dfrac{\ln 2}{\ln 4}\) is fine for A1.
dM1: Substitutes \(t = \ln 9\) and their evaluated \(k\) (which must be a numerical value) into one of their earlier equations connecting \(x\) and \(t\).
Note: that the 2nd Method mark is dependent on the 1st Method mark being awarded in part (c).
(corrected from the printed mark scheme: printed as “Substitutes \(t = \ln 4\) and their evaluated \(k\)”; the scheme substitutes \(t = \ln 9\).)
A1: \(x = \dfrac{2}{3}M\) cso.
Note: Please check their working as incorrect working can lead to a correct answer.